calorimeter's heat capacity?

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calorimeter's heat capacity?

A sample of octane is burned in a bomb calorimetergiving off 20.9 kJ of heat. The temperature of a 1155 gramquantity of water increases by 3.68oC. Calculate the calorimeter`s heat capacity.The specific heat of water is 4.184 J/goC

2007-03-15 19:15:13 · 2 answers · asked by ? 4 in Science & Mathematics ➔ Chemistry

2 answers

20.9 kJ = mcdT + CdT
20.9 kJ - mcdT = CdT
20.9 - 1155*4.184*3.68 = CdT
20.9-17.7837 =CdT
C=3.11633/dT
C=0.846828 J/'C

2007-03-15 19:22:12 · answer #1 · answered by jon r 1 · 1⤊ 0⤋

Let C be the calorimeter's heat capacity in J/oC

Heat liberated = heat gained to the water + heat gained by calorimeter. Thus

20900 = 1155 x 4.184 x 3.68 + 3.68 C leading to C = 847 J/oC

2007-03-15 19:51:54 · answer #2 · answered by A S 4 · 0⤊ 0⤋